What is the integral of sin^3(2θ)cos^{3/2}(2θ) ?

The integral of sin^3(2θ)cos^{3/2}(2θ) is -1/5 cos^{5/2}(2θ)+1/9 cos^{9/2}(2θ)+C

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integral of sin^3(2θ)cos^{3/2}(2θ)

\int sin^{3} (2 θ) cos^{\frac{3}{2}} (2 θ) d θ

- \frac{1}{5} cos^{\frac{5}{2}} (2 θ) + \frac{1}{9} cos^{\frac{9}{2}} (2 θ) + C

Solution steps

\int sin^{3} (2 θ) cos^{\frac{3}{2}} (2 θ) d θ

Apply u-substitution

: \int sin^{3} (u) cos^{\frac{3}{2}} (u) \frac{1}{2} d u

= \int sin^{3} (u) cos^{\frac{3}{2}} (u) \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sin^{3} (u) cos^{\frac{3}{2}} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{\frac{3}{2}} (u) d u

Apply u-substitution

: \int - v^{\frac{3}{2}} (1 - v^{2}) d v

= \frac{1}{2} \cdot \int - v^{\frac{3}{2}} (1 - v^{2}) d v

Expand

- v^{\frac{3}{2}} (1 - v^{2}) : - v^{\frac{3}{2}} + v^{\frac{7}{2}}

= \frac{1}{2} \cdot \int - v^{\frac{3}{2}} + v^{\frac{7}{2}} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int v^{\frac{3}{2}} d v + \int v^{\frac{7}{2}} d v)

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int v^{\frac{7}{2}} d v = \frac{2}{9} v^{\frac{9}{2}}

= \frac{1}{2} (- \frac{2}{5} v^{\frac{5}{2}} + \frac{2}{9} v^{\frac{9}{2}})

Substitute back

= \frac{1}{2} (- \frac{2}{5} cos^{\frac{5}{2}} (2 θ) + \frac{2}{9} cos^{\frac{9}{2}} (2 θ))

Simplify

\frac{1}{2} (- \frac{2}{5} cos^{\frac{5}{2}} (2 θ) + \frac{2}{9} cos^{\frac{9}{2}} (2 θ)) : - \frac{1}{5} cos^{\frac{5}{2}} (2 θ) + \frac{1}{9} cos^{\frac{9}{2}} (2 θ)

= - \frac{1}{5} cos^{\frac{5}{2}} (2 θ) + \frac{1}{9} cos^{\frac{9}{2}} (2 θ)

Add a constant to the solution

= - \frac{1}{5} cos^{\frac{5}{2}} (2 θ) + \frac{1}{9} cos^{\frac{9}{2}} (2 θ) + C

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What is the integral of sin^3(2θ)cos^{3/2}(2θ) ?
The integral of sin^3(2θ)cos^{3/2}(2θ) is -1/5 cos^{5/2}(2θ)+1/9 cos^{9/2}(2θ)+C